The reagent than can be used to distinguish between pent-1-yne and pent-2-yne is :

B r_{2} / C C l_{4}

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K M n O_{4} (a c i d i c)

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[A g (N H_{3})_{2}] N O_{3}

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O_{3} / H_{2} O

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Solution

The correct option is C $[A g (N H_{3})_{2}] N O_{3}$
Only terminal alkynes (pent-1-yne) react with ammoniacal solution of silver nitrate $(A g N O_{3})$ or cuprous chloride $(C u C l)$
$C H_{3} - C H_{2} - C H_{2} - C \equiv C H + 2 [A g (N H_{3})_{2}] N O_{3} \to C H_{3} - C H_{2} - C H_{2} - C \equiv C^{-} A g^{+} White precipitate$

$C H_{3} - C H_{2} - C \equiv C - C H_{3} + 2 [A g (N H_{3})_{2}] N O_{3} \to No reaction$

Hence option (c) is correct

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